# Geometric series algebra / exponential/ 2 summations

1. Nov 5, 2016

### binbagsss

The problem statement, all variables and given/known data

I want to show that $\sum\limits_{n=1}^{\infty} log (1-q^n) = -\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty} \frac{q^{n.m}}{m}$, where $q^{n}=e^{2\pi i n t}$ , $t$ [1] a complex number in the upper plane.

2. Relevant equations

Only that $e^{x} = \sum\limits_{m=0}^{\infty} \frac{x^{m}}{m!}$ [2]

3. The attempt at a solution

I can see that both series start from $n,m=1$ in the RHS of [1] , so if I use [2] starting from $m=1$ will result in the '1' cancelling as needed i.e. I have $\sum\limits_{n=1}^{\infty} log (1-q^n) = \sum\limits_{n=1}^{\infty} log (\sum\limits_{m=1}^{\infty} \frac{(2\pi int)^{m}}{m!})$

I don't really no where to go now, I don't see how you can get another expansion in terms of $q^{m}$ from this to give the required $q^{n.m}$, I can see there's a minus sign too but I'm struggling to use this as a clue as well.

2. Nov 5, 2016

### binbagsss

I am also struggling to show that $\sum\limits_{n=1}^{\infty} n \frac{q^{n}}{(1-q^n)} = \sum\limits_{n=1}^{\infty} n \sum\limits_{m=1}^{\infty} q^{m.n}$

I'm guessing it's the same things I'm missing in both cases...many thanks in advance.

3. Nov 5, 2016

### Fightfish

Unless I'm missing something, isn't this just a matter of Taylor expanding $\log(1-q^{n})$?